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Let $R$ be a ring and let $G$ be a group. I know that when $G \cong H \times K$, we have $RG \cong (RH)K$, meaning the group ring of $K$ over the ring $RH$. Is there anything similar when $R \cong S \times T$?

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Have you tried seeing if the obvious thing works? – Mariano Suárez-Alvarez♦ Jan 9 '12 at 3:18
@Mariano: I keep getting lost. – Nobody Jan 9 '12 at 3:28
Where? If you do not tell us what you tried, what you did and where you got stuck you leave us with no other option apart from doing it all for you. Moreover, there are great chances that the way we do it is not the same one you tried so reading what we would do will not do much to unstuck you from where you got stuck. (whathaveyoutried.com explains this more...) – Mariano Suárez-Alvarez♦ Jan 9 '12 at 3:31
Could you at least let me know what the obvious thing is and if it works? I'd be happy to work out the details myself. At least then I'd know I'm heading in the right direction. – Nobody Jan 9 '12 at 3:37
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A function from $G$ to $S\times T$ is the same as a pair of functions $G\to S$ and $G\to T$. This should suggest what the obvious thing to try is. – Rasmus Jan 9 '12 at 8:43

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